版权声明:本人版权意识薄弱,请随意转载 https://blog.csdn.net/Ike940067893/article/details/88775402
这个做法是
的主席树…
(其实还有不用主席树,只用分块+树状数组的
,因为常数小,快了一倍 戳这里)
CODE
貌似把块大小开成根号要12500ms,但是开成100只要9500ms…
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
template<typename T>inline void read(T &num) {
char ch; int flg = 1;
while((ch=getchar())<'0'||ch>'9')if(ch=='-')flg=-flg;
for(num=0;ch>='0'&&ch<='9';num=num*10+ch-'0',ch=getchar());
num*=flg;
}
const int MAXN = 50005;
int n, m, N, B, f[225][MAXN], a[MAXN], b[MAXN], bel[MAXN];
int ch[MAXN*20][2], sz[MAXN*20], rt[MAXN], tot;
void insert(int &i, int p, int l, int r, int x) {
sz[i=++tot] = sz[p] + 1;
if(l == r) return;
int mid = (l + r) >> 1;
if(x <= mid) ch[i][1] = ch[p][1], insert(ch[i][0], ch[p][0], l, mid, x);
else ch[i][0] = ch[p][0], insert(ch[i][1], ch[p][1], mid+1, r, x);
}
int query(int x, int y, int l, int r, int L, int R) {
if(sz[x] == sz[y]) return 0;
if(L <= l && r <= R) return sz[y]-sz[x];
int mid = (l + r) >> 1;
if(R <= mid) return query(ch[x][0], ch[y][0], l, mid, L, R);
else if(L > mid) return query(ch[x][1], ch[y][1], mid+1, r, L, R);
return query(ch[x][0], ch[y][0], l, mid, L, R) + query(ch[x][1], ch[y][1], mid+1, r, L, R);
}
int T[MAXN];
inline void upd(int x, int val) {
while(x) T[x] += val, x -= x&-x;
}
inline int qsum(int x) { int res = 0;
while(x <= N) res += T[x], x += x&-x;
return res;
}
inline int solve(int L, int R) {
if(L > R) swap(L, R);
int res = 0;
if(bel[L] == bel[R]) {
for(int i = L; i <= R; ++i)
res += qsum(a[i]+1), upd(a[i], 1);
for(int i = L; i <= R; ++i) upd(a[i], -1);
return res;
}
res = f[bel[L]+1][R];
for(int i = L; i <= bel[L]*B && i < R; ++i)
res += query(rt[i], rt[R], 1, N, 1, a[i]-1);
return res;
}
int main () {
read(n); B = sqrt(n);
for(int i = 1; i <= n; ++i) read(a[i]), b[++N] = a[i];
sort(b + 1, b + N + 1);
N = unique(b + 1, b + N + 1) - b - 1;
for(int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + N + 1, a[i]) - b;
for(int i = 1; i <= n; ++i) {
bel[i] = (i-1)/B + 1;
insert(rt[i], rt[i-1], 1, N, a[i]);
}
for(int i = 1; i <= bel[n]; ++i) {
for(int j = (i-1)*B+1; j <= n; ++j)
f[i][j] = f[i][j-1] + qsum(a[j]+1), upd(a[j], 1);
memset(T, 0, (N+1)<<2);
}
int ans = 0, x, y;
read(m);
while(m--) {
read(x), read(y), x ^= ans, y ^= ans;
printf("%d\n", ans=solve(x, y));
}
}